zoukankan      html  css  js  c++  java
  • POJ2533:Longest Ordered Subsequence

    Longest Ordered Subsequence


    Longest Ordered Subsequence
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 31680   Accepted: 13848

    Description

    A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

    Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

    Input

    The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

    Output

    Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

    Sample Input

    7
    1 7 3 5 9 4 8

    Sample Output

    4


    这是一道动归题。

    n=3


    1 7 3 5 9 4 8

    dp[]=0;

    1 ap[1]=0
    7 dp[2]=1
    3 dp[3]=1
    5 d[4]=2
    9 d[5]=3
    4 d[6]=2
    8 d[7]=3

    最后找出最大值然后加一;

    详细实现例如以下:


    
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    
    using namespace std;
    
    const int M = 1000+5;
    int sequence[M];    //输入的数据
    int dp[M];          
    
    int main()
    {
        int n;
        while(scanf("%d", &n)!=EOF)
        {
            for(int i=1; i<=n; i++)
                scanf("%d",  &sequence[i]);
            memset(dp, 0, sizeof(dp));          //初始化为0
            for(int i=1; i<=n; i++)
                for(int j=1; j<i; j++)
                {
                    if(sequence[i]>sequence[j])
                        dp[i] = max(dp[i], dp[j] + 1);    //动归方程
                }
            int ans=0;
            for(int i=1; i<=n; i++)
                ans = max(ans, dp[i]);
            printf("%d
    ", ans+1);
        }
        return 0;
    }


  • 相关阅读:
    oracle查看被锁的表和被锁的进程,杀掉进程
    umlの交互图
    Window XP安装Ubuntu14.04实现Samba文件共享
    开源企业IM免费企业即时通讯ENTBOOST V2014.177版本号正式公布
    必看的 jQuery性能优化的38个建议
    正则工具类以及FinalClass
    CF772E Verifying Kingdom
    aop相关术语
    BeanFactoryAware和BeanNameAware
    spring完成自动装配
  • 原文地址:https://www.cnblogs.com/mqxnongmin/p/10841138.html
Copyright © 2011-2022 走看看